# Some confusion about Gaussian ring [closed]

Is $$\mathbb{Z} [i]$$ is field ? yes/No

yes, I thinks it will field because it is integral domain

Is its True ?

## closed as off-topic by The Count, mrtaurho, Shailesh, Leucippus, ronnoJun 29 at 17:00

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• Try to divide $1$ by $2$. – darij grinberg Jun 29 at 14:37
• im not getting @darijgrinberg – jasmine Jun 29 at 14:38
• $1/2 \notin \mathbb{Z}[i]$ but if it were a field, $2$ would have to have a multiplicative inverse. – mathphys Jun 29 at 14:38
• $\mathbb Z$ is an integral domain. Is it a field? Why not? – lulu Jun 29 at 14:40
• Do you know the definitions of the words, "field", "integral domain"? I ask because you say you think it is a field because it is an integral domain. The set of all integers (with the usual addition and multiplication) is an integral domain (that's where the name comes from) but not a field. (Every field is an integral domain. Not every integral domain is a field.) – user247327 Jun 29 at 14:42

No, it is not a field. For instance, $$2$$ has no inverse in $$\mathbb Z[i]$$.
Well, $${\Bbb Z}[i] = \{a+bi\mid a,b\in{\Bbb Z}\}$$ is a subset of $${\Bbb C}=\{a+bi\mid a,b\in{\Bbb R}\}$$ and so an integral domain. But its clearly not a field.
Note that $$\mathbb{Z} \subseteq \mathbb{Z}[i]$$ and $$\mathbb{Z}$$ is not a field. To see this, we know that $$1$$ is the identity in $$\mathbb{Z}$$, so how would we invert $$2$$? We would need to multiply it by $$\frac{1}{2}$$, which is not an element of $$\mathbb{Z}$$. In fact, the only invertible elements in $$\mathbb{Z}$$ are $$\pm 1$$, hence the integers do not form a field.
Hence, $$\mathbb{Z}[i]$$ is not a field.
We know that $$\mathbb Z[i]\cong\mathbb Z[x]/(x^2+1).$$
And $$(x^2+1)\subsetneq (3,x^2+1)\subsetneq\mathbb Z[x]\quad \text{(why?)}$$ implies $$(x^2+1)$$ is not a maximal ideal of $$\mathbb Z[x]$$, hence $$\mathbb Z[i]$$ is not a field.